I mean swapItems(), removeItem(), addItem(), setItem() ?? What is the purpose of this D-shaped ring at the base of the tongue on my hiking boots? Find centralized, trusted content and collaborate around the technologies you use most. Sorting list according to corresponding values from a parallel list What do you mean when you say that you're unable to persist the order "on the backend"? Sorting for String values differs from Integer values. This is just an example, but it demonstrates an order that is defined by a list, and not the natural order of the datatype: Now, let's say that listA needs to be sorted according to this ordering. 2023 DigitalOcean, LLC. How can we prove that the supernatural or paranormal doesn't exist? I am wondering if there is any easier way to do it. Sign up for Infrastructure as a Newsletter. I think that the title of the original question is not accurate. Application of Binary Tree. We first get the String values in a list. Let's start with two entity classes - Employee and Department: class Employee { Integer employeeId; String employeeName; // getters and setters } class Department { Integer . Once you have that, define your own comparison function which compares values based on the indexes of list. 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Site design / logo 2023 Stack Exchange Inc; user contributions licensed under CC BY-SA. Sorting Strings in reverse order is as simple as sorting integers in reverse order: In all of the previous examples, we've worked with Comparable types. But it should be: The list is ordered regarding the first element of the pairs, and the comprehension extracts the 'second' element of the pairs. The signature of the method is: Let's see another example of Collections.sorts() method. Merge two lists in Java and sort them using Object property and another condition, How Intuit democratizes AI development across teams through reusability. I want to sort listA based on listB. If they are already numpy arrays, then it's simply. The signature of the method is: T: Comparable type of element to be compared. Note also, that the SortedDependingList does currently not allow to add an element from listA a second time - in this respect it actually works like a set of elements from listA because this is usually what you want in such a setting. We can also create a custom comparator to sort the hash map according to values. A:[c,b,a] Else, run a loop till the last node (i.e. Why are physically impossible and logically impossible concepts considered separate in terms of probability? You can use a Bean Comparator to sort this List however you desire. If you already have a dfwhy converting it to a list, process it, then convert to df again? Warning: If you run it with empty lists it crashes. My solution: The time complexity is O(N * Log(N)). No new elements. Here is Whatangs answer if you want to get both sorted lists (python3). Thanks for learning with the DigitalOcean Community. The most obvious solution to me is to use the key keyword arg. Assuming that the larger list contains all values in the smaller list, it can be done. There is a difference between the two: a class is Comparable when it can compare itself to another class of the same type, which is what you are doing here: one Factory is comparing itself to another object. It seems what you want would be to use Comparable instead, but even this isn't a good idea in this case. That is, the first items (from Y) are compared; and if they are the same then the second items (from X) are compared, and so on. 2. @Richard: the keys are computed once before sorting; so the complexity is actually O(N^2). It is from Java 8. For more information on how to set\use the key parameter as well as the sorted function in general, take a look at this. [[name=a, age=age11], [name=a, age=age111], [name=a, age=age1], [name=b, age=age22], [name=b, age=age2], [name=c, age=age33], [name=c, age=age3]]. @Hatefiend interesting, could you point to a reference on how to achieve that? Are there tables of wastage rates for different fruit and veg? I like having a list of sorted indices. The second one is easier and faster if you're not using Pandas in your program. @RichieV I recommend using Quicksort or an in-place merge sort implementation. Just remember Zx and Zy are tuples. But it should be: The list is ordered regarding the first element of the pairs, and the comprehension extracts the 'second' element of the pairs. Here if the data type of Value is String, then we sort the list using a comparator. It is the method of Java Collections class which belong to a java.lang package. That's easily managed with an index list: Since the decorate-sort-undecorate approach described by Whatang is a little simpler and works in all cases, it's probably better most of the time. Not the answer you're looking for? Sorry, that was my typo. Best answer! If you have 2 lists of identical number of items and where every item in list 1 is related to list 2 in the same order (e.g a = 0 , b = 1, etc.) How Intuit democratizes AI development across teams through reusability. It only takes a minute to sign up. Code Review Stack Exchange is a question and answer site for peer programmer code reviews. How can this new ban on drag possibly be considered constitutional? Sorting a list based on another list's values - Java 16,973 Solution 1 Get rid of the two Lists. When we try to use sort over a zip object. You can create a pandas Series, using the primary list as data and the other list as index, and then just sort by the index: This is helpful when needing to order a smaller list to values in larger. The end result should be list Y being untouched and list X being changed into the expected solution without ever having to create a temp list. Whats the grammar of "For those whose stories they are"? Let's say we have the following code: Let's sort them by age, first. zip, sort by the second column, return the first column. If you try your proposed code, it would give something like this: Person{name=Giant L2, age=100} Person{name=Derp L1, age=50} Person{name=John L2, age=50} Person{name=Menard L1, age=44} Person{name=Lili L1, age=44} Person{name=Lili L2, age=44} Person{name=Menard L2, age=44} Person{name=Bob L1, age=22} Person{name=Alec L1, age=21} Person{name=Herp L1, age=21} Person{name=Alec L2, age=21} Person{name=Herp L2, age=21} Person{name=Alice L1, age=12} Person{name=Little L2, age=5} And it's not what I'm looking for. We can also pass a Comparator implementation to define the sorting rules. "After the incident", I started to be more careful not to trip over things. How to remove an element from a list by index, Sorting an array of objects by property values, String formatting: % vs. .format vs. f-string literal. As for won't work..that's right because he posted the wrong question in the title when he talked about lists. By default, the sort () method sorts a given list into ascending order (or natural order ). Any suggestions? That is, the first items (from Y) are compared; and if they are the same then the second items (from X) are compared, and so on. More general case (sort list Y by any key instead of the default order), http://scienceoss.com/sort-one-list-by-another-list/, How Intuit democratizes AI development across teams through reusability. Surly Straggler vs. other types of steel frames. Merge two lists in Java and sort them using Object property and another Most of the following examples will use lists but the same concept can be applied for arrays. Once we have the list of values in a sorted manner, we build the HashMap again based on this new list. I am a bit confused with FactoryPriceComparator class. @RichieV I recommend using Quicksort or an in-place merge sort implementation. Premium CPU-Optimized Droplets are now available. A Comparator can be passed to Collections.sort () or List.sort () method to allow control over the sort order. Once you have a list of sorted indices, a simple list comprehension will do the trick: Note that the sorted index list can also be gotten using numpy.argsort(). Something like this? How do I sort a list of dictionaries by a value of the dictionary? QED. In Java How to Sort One List Based on Another. In this tutorial, we will learn how to sort a list in the natural order. This method returns a lexicographic-order comparator with another comparator. In Java how do you sort one list based on another?